$h(t) = \begin{cases} \dfrac{t}{3}+6&, & t\in (-\infty,-3) \\\\ t(t+4) &, & t \in [-3,3)\\\\ t^2+t+1 &, & t \in [3,\infty)\end{cases}$ $h(0)=$
Answer: The strategy First, we should find the appropriate assignment rule out of the three, by checking which case applies for $t={0}$. [I don't understand the notation for the cases.] Finding the appropriate assignment rule Since ${0}\in[-3,3)$, we should use the second assignment rule $t(t+4)$. The answer $h({0})={0}({0}+4)=0$ In conclusion, $h(0)=0$.